There is multiple way to calculate a square root https://en.wikipedia.org/wiki/Methods_of_computing_square_roots The simples to understand is iterative methods. The general idea is make a guess and then improve it.

without using any specific algorithm let's try to calculate the square root of 26. Let's find a to large and to small number and test the average of them. If the average is to small replace the smaller number, if it is to larger replace the larger

Let's first guess at 5 5 ^(2) =25, that is too low lets call it S

The second guess at 6 6^(2) =36 is to large let's call it L .

Let's try the midpoint (S+ L)/2 =(5+6)/2 = 5.5 and 5.5^2 =30.25 which is to large so replace S with it.

The next average is (5+5.5)/2=5.25 Let's repeat it in the table below. I rounded all numbers to 6 or less decimals

S        L        (S+L)/2        ((S+L)/2)^2
5        6           5.5            30.25        To large
5        5.5         5.25           27.5625      To large
5        5.25        5.125          26.265625    To large
5        5.125       5.0625         25.628906    To small
5.0625   5.125       5.09375        25.946289    To small
5.09375  5.125       5.109375       26.105716    To large
5.09375  5.109375    5.101563       26.02594     To large
5.09375  5.101563    5.097659       25.986102    To small
5.097659 5.101563    5.099611       26.006032    To large
5.097659 5.099611    5.098635       25.996079    To small
5.098635 5.099611    5.099123       26.001055    To large
5.098635 5.099123

Let's stop at this point. We can see that the square root is in between 5.098635 and 5.099123 so we know it to be two decimals as 5.09. We could continue this forever and get closer and closer. Many roots like this will have an infinite number of decimals so wee needs to stop.

The algorithm above is not an especially efficient one because it converges quite slowly but it is a good example of an iterative way do numerical find a solution to an equation.

The guesses I made were so you quickly get some decimals correctly. But you can pick a conservative guess like 1 and the number /2 that will work for any number we look for a root for that is larger than 1

Newton-Rapson https://en.wikipedia.org/wiki/Newton%27s_method method is faster, but exactly why you choose that formulais a lot harder to explain.

Let's call x the guess and a number we look for the root of in this case a=26. The next guess can be calculated 1/2( x + a/x) let's start with 5 and calculate the next number as 1/2 (5+26/5) = 5.1 . The next step is 1/2 (5.1+26/5.1)

I added a space among the decimal at the point it is longer accurate

5
5.1
5.099019 6078431372549019607843137254901960784313725490196078431372
5.09901951359278 57010906650681807043140270913210506275188406452756
5.099019513592784830028224109022 8563911005788636011794938296893309
5.099019513592784830028224109022781989563770946099596407584970804 9


The square root is
5.0990195135927848300282241090227819895637709460995964075849708044...

The number of accurate decimals doubles in each step. That is a very efficient way to find a square root of a number